combustion of hydrocarbons such as octane (c₈h₁₈) produces carbon dioxide, a \greenhouse gas.\ greenhouse…

combustion of hydrocarbons such as octane (c₈h₁₈) produces carbon dioxide, a \greenhouse gas.\ greenhouse gases in the earths atmosphere can trap the suns heat, raising the average temperature of the earth. for this reason there has been a great deal of international discussion about whether to regulate the production of carbon dioxide.\n1. write a balanced chemical equation, including physical state symbols, for the combustion of liquid octane into gaseous carbon dioxide and gaseous water.\n2. suppose 0.330 kg of octane are burned in air at a pressure of exactly 1 atm and a temperature of 20.0 °c. calculate the volume of carbon dioxide gas that is produced. be sure your answer has the correct number of significant digits.

combustion of hydrocarbons such as octane (c₈h₁₈) produces carbon dioxide, a \greenhouse gas.\ greenhouse gases in the earths atmosphere can trap the suns heat, raising the average temperature of the earth. for this reason there has been a great deal of international discussion about whether to regulate the production of carbon dioxide.\n1. write a balanced chemical equation, including physical state symbols, for the combustion of liquid octane into gaseous carbon dioxide and gaseous water.\n2. suppose 0.330 kg of octane are burned in air at a pressure of exactly 1 atm and a temperature of 20.0 °c. calculate the volume of carbon dioxide gas that is produced. be sure your answer has the correct number of significant digits.

Answer

Explanation:

Step1: Write balanced chemical equation

The combustion of liquid octane ($C_8H_{18}(l)$) in oxygen ($O_2(g)$) to form gaseous carbon - dioxide ($CO_2(g)$) and gaseous water ($H_2O(g)$) is: $$2C_8H_{18}(l)+25O_2(g)\rightarrow16CO_2(g) + 18H_2O(g)$$

Step2: Calculate moles of octane

The molar mass of octane ($C_8H_{18}$) is $M=(8\times12.01 + 18\times1.01)\ g/mol=114.23\ g/mol$. The mass of octane is $m = 0.330\ kg=330\ g$. The number of moles of octane, $n_{C_8H_{18}}=\frac{m}{M}=\frac{330\ g}{114.23\ g/mol}\approx2.89\ mol$.

Step3: Determine moles of carbon - dioxide

From the balanced equation, the mole ratio of $C_8H_{18}$ to $CO_2$ is $2:16$ or $1:8$. So, $n_{CO_2}=8\times n_{C_8H_{18}}$. $n_{CO_2}=8\times2.89\ mol = 23.12\ mol$.

Step4: Use ideal gas law to find volume of carbon - dioxide

The ideal gas law is $PV = nRT$, where $P = 1\ atm$, $n=n_{CO_2}=23.12\ mol$, $R = 0.0821\ L\cdot atm/(mol\cdot K)$ and $T=(20.0 + 273.15)\ K=293.15\ K$. We solve for $V$: $V=\frac{nRT}{P}$. $V=\frac{23.12\ mol\times0.0821\ L\cdot atm/(mol\cdot K)\times293.15\ K}{1\ atm}\approx555\ L$.

Answer:

555